Problem: A rectangular prism measures 10-inches by 20-inches by 10-inches. What is the length, in inches, of the diagonal connecting point A and point B? Express your answer in simplest radical form. [asy]
unitsize(0.75cm);
defaultpen(linewidth(0.7pt)+fontsize(10pt));
dotfactor=4;

draw((0,1)--(1,1)--(1,0)--(0,0)--(0,1)--(1,2)--(2,2)--(1,1));
draw((1,0)--(2,1)--(2,2));
dot((0,1));
label("$A$",(0,1),W);
dot((2,1));
label("$B$",(2,1),E);
[/asy]
Answer: Let $D$ be the vertex opposite $B$ on the bottom face, and let $C$ be one of the other two vertices on the bottom face.  Because $BCD$ is a right triangle, we can use the Pythagorean theorem to get $BD=\sqrt{10^2+20^2}$.  Then applying the Pythagorean theorem to right triangle $BDA$, we find \begin{align*} AB&=\sqrt{10^2+\left(\sqrt{10^2+20^2}\right)^2}\\&=\sqrt{10^2+10^2+20^2}\\&=\sqrt{600}=\boxed{10\sqrt{6}} \text{ inches}.\end{align*}

[asy]
size(150);
defaultpen(linewidth(0.7pt)+fontsize(10pt));
dotfactor=4;
draw((0,1)--(1,1)--(1,0)--(0,0)--(0,1)--(1,2)--(2,2)--(1,1));
draw((1,0)--(2,1)--(2,2));
dot((1,0));
label("$C$",(1,0),SE);
dot((0,0));
label("$D$",(0,0),SW);
dot((0,1));
label("$A$",(0,1),W);
dot((2,1));
label("$B$",(2,1),E);
[/asy]